Part 21

Set revolutions on B to horse-power on A, and bring cursor to 50 on B. Then move the slide until the same number is found on B under the cursor that is simultaneously found on D under the index of C. This number is the diameter required.

To find the deflection k in inches, of a round steel shaft of diameter d, under a uniformly distributed load in lb. w, and supported by bearings, the centres of which are l feet apart (k = (w l^3)/(78,000d^4)).

Modifying the form of this expression slightly, we proceed as follows:—Set d on C to l on D, and bring the cursor to the same number on B that is found on D under the index of C. Bring d on B to cursor, cursor to w on B, 78,800 on B to cursor, and read deflection on A over index of B.

EX.—Find the deflection in inches of a round steel shaft 3½in. diameter, carrying a uniformly distributed load of 3200 lb., the distance apart of the centres of support being 9 ft.

Set 3·5 on C to 9 on D, and read 2·57 on D, under the L.H. index of C. Set cursor to 2·57 on B, and bring 3·5 on B to cursor, cursor to 3200 on B, 78,000 on B to cursor, and over L.H. index of B read 0·199 in., the required deflection on A.

To find the diameter of a shaft subject to twisting only, given the twisting moment in inch-lb. and the allowable stress in lb. per square inch.

Set the stress in lb. per square inch on B to the twisting moment in inch-lb. on A, and bring cursor to 5·1 on B. Then move the slide until the same number is found on B under the cursor that is simultaneously found on D under the index of C.

EX.—A steel shaft is subjected to a twisting moment of 2,700,000 inch-lb. Determine the diameter if the allowable stress is taken at 9000 lb. per square inch.

Set 9000 on B to 2,700,000 on A, and bring the cursor to 5·1 on B. Moving the slide to the left, it is found that when 11·51 on the R.H. scale of B is under the cursor, the L.H. index of C is opposite 11·51 on D. This, then, is the required diameter of the shaft.

(N.B.—The rules for the scales to be used in finding the cube root (page 42) must be carefully observed in working these examples.)

MOMENTS OF INERTIA.

To find the moment of inertia of a square section about an axis formed by one of its diagonals (I = (s^4)/(12)).

Set index of C to the length of the side of square s on D; bring cursor to s on C, 12 on B to cursor, and over index of B read moment of inertia on A.

To find the moment of inertia of a rectangular section about an axis parallel to one side and perpendicular to the plane of bending.

Set index of C to the height or depth h of the section, and bring cursor to h on B. Set 12 on B to cursor, and over breadth b of the section on B read moment of inertia on A.

EX.—Find the moment of inertia of a rectangular section of which h = 14 in. and b = 7 in.

Set index of C to 14 on D, and cursor to 14 on B. Bring 12 on B to cursor, and over 7 on B read 1600 on A.

DISCHARGE FROM PUMPS, PIPES, ETC.

To find the theoretical delivery of pumps, in gallons per stroke.

Set 29·4 on B to the diameter of the plunger in inches on D, and over length of stroke in feet on B read theoretical delivery in gallons per stroke on A.

(N.B.—A deduction of from 20 to 40 per cent. should be made to allow for slip.)

To find loss of head of water in feet due to friction in pipes (Prony’s rule).

Set diameter of pipe in feet on B to velocity of water in feet per second on D and bring cursor to 2·25 on B; bring 1 on B to cursor, and over length of pipe in miles on B, read loss of head of water in feet, on A.

To find velocity in feet per second, of water in pipes (Blackwell’s rule).

Set 2·3 on B to diameter of pipe in feet on A, and under inclination of pipe in feet per mile on B read velocity in feet per second on D.

To find the discharge over weirs in cubic feet per minute and per foot of width. (Discharge = 214√(h^3))

Set 0·00467 on C to the head in feet h on D, and under h on B read discharge on D.

To find the theoretical velocity of water flowing under a given head in feet.